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Full text of "Treatise On Analysis Vol-Ii"

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But, by virtue of (, (V- ng(x)\ng(yj) = 0 for all x9 y in A. Since
A/U0 is dense in Hg , it follows that V= 0 and hence F = 0.              Q.E.D.


(15.9.5)    We have seen ( that a positive linear form /on an algebra
with involution A having a unit element always satisfies the condition (B). We
recall that if A is in addition a Banach algebra, then the corresponding Hilbert
form g also satisfies (U) (15.6.11). If A is a separable Banach algebra with
unit element e, then the prehilbert space A/ng and the star algebra jtfg are also
separable. The first assertion follows from the continuity of/ (15.6.11), which
implies that \\ng(x)\\2 =/(x*jc)  ||/||  \\x*x\\  \\f\\  ||x||2 and shows there-
fore that the image under ng of a denumerable dense set in A is dense in the
prehilbert space A/ng . The separability of jtfg follows from the fact that the
representation Ug is continuous (15.5.7) and therefore transforms a denumer-
able dense set in A into a denumerable dense set in jfg . Hence the Bochner-
Godement theorem can be applied to every positive linear form on A.

(15.9.6)    If s\-> U(s) is a representation of A on a Hilbert space H, then for
each x0 E H the form fXQ(s) = (U(s)  x0 \XQ) satisfies the condition (B),
because by (6.2.4)

\fxo(s)\2 g ||E/(s)  x0f  IK!]2 = \\x0[\2fxo(s*s).

By (15.5.6) this shows that knowledge of the hermitian characters of a
commutative algebra A with involution which has a unit element determines
all the representations of A. In addition, we have seen in the course of the
proof of (15.9.4) that, if/ is a positive linear form satisfying the conditions
of (15.9.4(ii)), then the corresponding representation x\-+Ug(x) admits a
totalizer a (


(15.9.7) It can happen that the formula (15.9.3) (and the conditions of
(15.9.2)) are satisfied but that the measure mg is not bounded (and therefore
/does not satisfy (B)) (Problem 2). On the other hand, there are examples
in which the conditions (U) and (N) are satisfied but (B) is not (Problem 4),
and examples in which (B) and (N) are satisfied but (U) is not (Problem 5).ly on S^ to G, and hence,